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10
7
steel
(
E
=
3
×
psi
).
/
∂
P
Internal
Axial Force in Each
Member Due to
∂
N
Internal Axial
a Unit Load at
b
Force
N
in Each
in
th
e Direction
Member Due to
of
P
. All Actual
Length
Area
the Actual
L
oads
Applied Loads are
N
∂
N
∂
P
L
EA
Member
L
(in.)
A
(in.
2
)
(here
P
)
Removed
100
√
2
5000
√
2
−
√
2
5
√
2
1
2
−
/
2
(
10
−
2
)
/
6
/
(
10
−
2
)
/
2
100
1
5000
1
2
5
6
10
−
1
3
100
1
10,000
1
(
)
1
/
3
4
100
1
5000
1
/
2
(
10
−
2
)
5
/
6
100
√
2
5000
√
2
−
√
2
5
√
2
10
−
2
5
2
−
/
2
(
)
/
6
N
∂
N
∂
P
L
EA
V
b
=
=
0
.
00736 in
.
5 bars
Although the use of the unit load is recommended in most cases,
fo
r this simple truss
th
e
N
column could just as well have been ex
pre
ssed in terms of
P
(rather than using
P
=
∂
/∂
10,000) and then differentiated to find
N
P
directly. These entries would then have
appeared as
Member
N
∂
N
/∂
P
P
√
2
−
√
2
1
−
/
2
/
2
2
P
/
2
1
/
2
3
P
1
4
P
/
2
1
/
2
P
√
2
−
√
2
5
−
/
2
/
2
EXAMPLE 3.8 Frame Analysis
Compute the reactions of the statically indeterminate frame of Fig. 3.5a. Consider only the
effects of bending in each of the members which form the frame.
Use Castigliano's theorem, part II in the form
U
∗
=
∂
1
EI
∂
M
u
d
R
Vd
=
0
=
M
dx
∂
∂
R
Vd
U
∗
w
d
=
∂
1
EI
∂
M
R
Hd
=
0
=
M
dx
(1)
∂
∂
R
Hd
U
∗
=
∂
1
EI
M
∂
M
θ
M
d
=
0
=
dx
d
∂
∂
M
d
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